Post on 21-Jul-2016
Chapter 3: Calculations Chapter 3: Calculations and the Chemical Equationand the Chemical Equation
The Mole Concept and Atoms
Atomic mass unitAtomic mass unit
1 amu = 1.661 X 101 amu = 1.661 X 10-24-24 g g
Because the mass of one amu is so small, Because the mass of one amu is so small, chemists deal with a much larger number chemists deal with a much larger number of atoms while working with chemicalsof atoms while working with chemicals
Mole Mole
One mole is defined as 6.022 X 10One mole is defined as 6.022 X 102323. . This refers to one mole of anything, eggs, This refers to one mole of anything, eggs, paperclips, atoms. One mole of anything paperclips, atoms. One mole of anything is 6.022 X 10is 6.022 X 1023 23 items. Much like one items. Much like one dozen of something is 12.dozen of something is 12.This number, 6.022 X 10This number, 6.022 X 102323 is called is called Avogadro’s number, named after the Avogadro’s number, named after the scientist who conducted a series of scientist who conducted a series of experiments leading to the “mole concept”.experiments leading to the “mole concept”.
The mole conceptThe mole concept
The mole and the amu are related. For The mole and the amu are related. For atoms, the atomic mass of an element atoms, the atomic mass of an element corresponds to the average mass of a corresponds to the average mass of a single atom in amu single atom in amu
AndAnd
The mass of a mole of atoms in grams.The mass of a mole of atoms in grams.
For example:For example:
The atomic mass of oxygen is 16.00 amu.The atomic mass of oxygen is 16.00 amu.
AndAnd
One mole of oxygen atoms (6.022 X 10One mole of oxygen atoms (6.022 X 102323 oxygen atoms) has a mass of 16.00 gramsoxygen atoms) has a mass of 16.00 grams
Another exampleAnother example
The atomic mass of iron (Fe) is 55.85 The atomic mass of iron (Fe) is 55.85 amu.amu.
AndAnd
One mole of iron atoms (6.022 X 10One mole of iron atoms (6.022 X 102323 oxygen atoms) has a mass of 55.85 gramsoxygen atoms) has a mass of 55.85 grams
And yet another exampleAnd yet another example
The atomic mass of radium (Ra) is 226 The atomic mass of radium (Ra) is 226 amu.amu.
AndAnd
One mole of radium atoms (6.022 X 10One mole of radium atoms (6.022 X 102323 radium atoms) has a mass of 226 gramsradium atoms) has a mass of 226 grams
Molar massMolar mass
The mass of one mol (mole) of atoms in The mass of one mol (mole) of atoms in gramsgrams
NoteNote
One mole of atoms of any element One mole of atoms of any element contains 6.022 X 10contains 6.022 X 102323 atoms, regardless of atoms, regardless of the type of element.the type of element.The mass of one mole of an element The mass of one mole of an element depends on what that element is, and is depends on what that element is, and is equal to the atom mass of that element in equal to the atom mass of that element in grams.grams.
This meansThis means
There are several “conversions” regarding There are several “conversions” regarding atoms, moles, and massatoms, moles, and mass
Converting moles to atomsConverting moles to atoms
How many atoms are in 4 moles of H?How many atoms are in 4 moles of H?
4 moles H X 6.022 X 104 moles H X 6.022 X 102323 atoms/mole = atoms/mole = 24.088 X 1024.088 X 102323 atoms of hydrogen or 2.409 atoms of hydrogen or 2.409
X 10X 102424 atoms atoms
In this case you multiply the number of In this case you multiply the number of moles X the number of atoms in each moles X the number of atoms in each mole.mole.
Converting atoms to molesConverting atoms to molesCalculate the number of moles of copper represented by Calculate the number of moles of copper represented by 3.26 X 103.26 X 102424 atoms. atoms.
3.26 X 103.26 X 1024 24 = 32.6 X 10 = 32.6 X 102323 (ok, I did this step to make the (ok, I did this step to make the math easier.)math easier.)
32.6 X 1032.6 X 102323 / 6.022 X 10 / 6.022 X 102323 = 5.413 X 10 = 5.413 X 102323 moles of copper. moles of copper.
In this case, to convert atoms to moles, I divide the In this case, to convert atoms to moles, I divide the number of atoms by the number of atoms in one mol (by number of atoms by the number of atoms in one mol (by 6.022 X 106.022 X 102323 ) )
Converting moles of a substance to Converting moles of a substance to mass in grams.mass in grams.
What is the mass in grams of 5.6 mol of What is the mass in grams of 5.6 mol of Neon?Neon?
The mass of one mole of Ne is the same The mass of one mole of Ne is the same as the atomic mass in g (20.18 g)as the atomic mass in g (20.18 g)
5.6 mol X 20.18 g/mol = 100.9 g of Ne5.6 mol X 20.18 g/mol = 100.9 g of Ne
Converting grams to numbers of Converting grams to numbers of atoms.atoms.
How many atoms would be in a gold ring How many atoms would be in a gold ring that weighs 25 g?that weighs 25 g?First, find the number of moles of Gold in First, find the number of moles of Gold in 25 g. Gold has an atomic mass of 107.9. 25 g. Gold has an atomic mass of 107.9. So, 25 g / 107.9 g/mol = 0.2317 mol of So, 25 g / 107.9 g/mol = 0.2317 mol of gold are in the ring.gold are in the ring.Next, 0.2317 mol X (6.022 X 10Next, 0.2317 mol X (6.022 X 102323) ) atoms/mol =1.395 X 10atoms/mol =1.395 X 1023 23 atomsatoms
When dealing with molecules. . .When dealing with molecules. . .
Like OLike O2 2 or Hor H2 2 , double the number of atoms,
because there are 2 atoms per molecule.Remember, one mole of something is 6.022 X X 101023 23 of whatever it is. If it is molecules, it’sof whatever it is. If it is molecules, it’s 6.022 X 10 1023 23 of themof them.. If it is atoms, it’s 6.022 x10 If it is atoms, it’s 6.022 x1023 23
atoms. atoms. If there are 2 atoms per molecule you need to If there are 2 atoms per molecule you need to double the number of moles. double the number of moles. 2 X (2 X (6.022 X 10X 1023 23 ) = 12.044 X 10) = 12.044 X 1023 23 or 1.204 X or 1.204 X 10102424
Homework Assignment # 10Homework Assignment # 10
Read p. 119-123. Read p. 119-123. As you read, complete exercises 1-6. As you read, complete exercises 1-6. When you are done reading, answer When you are done reading, answer problems 23-36 on p. 146-147problems 23-36 on p. 146-147
Chapter 4: Calculations Chapter 4: Calculations and the Chemical Equationand the Chemical Equation
Section 4.2: The Chemical Section 4.2: The Chemical Formula, Formula Weight, and Formula, Formula Weight, and
Molar MassMolar Mass
Chemical FormulaChemical Formula
A combination of symbols of the various A combination of symbols of the various elements that make up the compound.elements that make up the compound.
Formula UnitFormula Unit
The smallest amount of atoms that The smallest amount of atoms that provides the following informationprovides the following information The identity of atoms in the compoundThe identity of atoms in the compound The relative numbers of each type of atomThe relative numbers of each type of atom ExamplesExamples
Molecule vs ion pairMolecule vs ion pair
Covalent compounds form molecules, and Covalent compounds form molecules, and when calculating formula weight all of the when calculating formula weight all of the atoms in the compound are added atoms in the compound are added together.together.Ion pairs (ionic compounds) form Ion pairs (ionic compounds) form crystalline structures. It’s the smallest crystalline structures. It’s the smallest group of ions that are listed in the formula group of ions that are listed in the formula for these types of chemicals.for these types of chemicals.
Formula weight vs Molecular Formula weight vs Molecular WeightWeight
The sum of all of the atomic weights in the The sum of all of the atomic weights in the compound in an ionic compound it’s the compound in an ionic compound it’s the formula weight. In a covalent compound formula weight. In a covalent compound it’s the molecular weight.it’s the molecular weight.
Molar MassMolar Mass
The mass of one mole of the compound or The mass of one mole of the compound or the formula weight in grams.the formula weight in grams.ExamplesExamples
Conversions using Formula WeightConversions using Formula Weight
Finding the number of moles Finding the number of moles corresponding to a certain number of corresponding to a certain number of grams.grams.
Conversions using Formula WeightConversions using Formula Weight
Finding grams corresponding to a certain Finding grams corresponding to a certain number of moles.number of moles.
Homework Assignment #11Homework Assignment #11
Read p. 123-126.Read p. 123-126.On p. 147 Exercises 37-58On p. 147 Exercises 37-58
Chapter 4: Calculations Chapter 4: Calculations and the Chemical Equationand the Chemical Equation
Section 3: The Chemical Section 3: The Chemical Equation and the Information it Equation and the Information it
ConveysConveys
Chemical equationChemical equation
The shorthand notation for a chemical The shorthand notation for a chemical reaction, where one substance changes reaction, where one substance changes chemically into another substance.chemically into another substance.An example: burning sugarAn example: burning sugar
ReactantsReactants
The starting materials that undergo a The starting materials that undergo a chemical changechemical change
ProductsProducts
The ending materials that are produced by The ending materials that are produced by a chemical reaction.a chemical reaction.
Additional information in a chemical Additional information in a chemical reactionreaction
Physical state of the substance (solid, Physical state of the substance (solid, liquid, or gas)liquid, or gas)If the reaction occursIf the reaction occursIdentifies the solvent, if there is one. (A Identifies the solvent, if there is one. (A solvent is the solution the materials are solvent is the solution the materials are dissolved in, such as water.)dissolved in, such as water.)Experimental conditions such as heat, Experimental conditions such as heat, light, or electrical energy addedlight, or electrical energy added
Most importantlyMost importantly
The chemical equation identifies the The chemical equation identifies the relative number of moles of reactants and relative number of moles of reactants and products.products.
According to the Law of According to the Law of Conservation of MassConservation of Mass
Matter cannot be gained or lost in the Matter cannot be gained or lost in the process of a chemical reactionprocess of a chemical reactionThe total mass of the products must equal The total mass of the products must equal the total mass of the reactantsthe total mass of the reactantsThe chemical equation must be balanced.The chemical equation must be balanced.
Features of a chemical reaction.Features of a chemical reaction.
CaCOCaCO33(s) (s) →→∆ ∆ CaO(s) + COCaO(s) + CO22(g)(g)
Reactants are on the left of the
arrow.
The arrow indicates the
reaction occurs in one
direction.
The products
are on the right of the
arrow.
Features of a chemical reaction.Features of a chemical reaction.
CaCOCaCO33(s) (s) →→∆ ∆ CaO(s) + COCaO(s) + CO22(g)(g)
“s” indicates the chemical
is a solid substance
“g” indicates the substance
is a gas
“l” would indicate the
substance
were a liquid.
Features of a chemical reaction.Features of a chemical reaction.
CaCOCaCO33(s) (s) →→∆ ∆ CaO(s) + COCaO(s) + CO22(g)(g)
The ∆ indicates that energy was necessary for the chemical reaction to
occur
Features of a chemical reaction.Features of a chemical reaction.
CaCOCaCO33(s) (s) →→∆ ∆ CaO(s) + COCaO(s) + CO22(g)(g)
The main feature of a chemical equation is that it is balanced, with the same number of elements in
compounds on both sides of the arrow.
The experimental basis of a The experimental basis of a chemical equationchemical equation
Evidence for a chemical reaction includes:Evidence for a chemical reaction includes:
The release of a gas resulting in bubblesThe release of a gas resulting in bubbles
The formation of a solid (precipitate) in solutionThe formation of a solid (precipitate) in solution
The production of heat resulting in an increase in The production of heat resulting in an increase in temperaturetemperature
A change in color of a substanceA change in color of a substance
The experimental basis of a The experimental basis of a chemical equationchemical equation
Sometimes instruments must be used to Sometimes instruments must be used to measure subtle changes that indicate a measure subtle changes that indicate a chemical reaction.chemical reaction.
Heat or light absorbed or emittedHeat or light absorbed or emitted
Changes in the way a sample behaves in an electrical Changes in the way a sample behaves in an electrical or magnetic fieldor magnetic field
Changes in electrical propertiesChanges in electrical properties
Writing Chemical ReactionsWriting Chemical Reactions
Most reactions follow a few simple Most reactions follow a few simple patternspatterns
Combination reactionsCombination reactions
Decomposition reactionsDecomposition reactions
Replacement reactionsReplacement reactions
Combination reactionsCombination reactions
Involve the joining or combining of two or Involve the joining or combining of two or more compoundsmore compoundsThe general form of the reaction is The general form of the reaction is
A + B A + B → AB→ AB
Combination reactionsCombination reactions
Examples include:Examples include: Combination of a metal and non-metal to form a saltCombination of a metal and non-metal to form a salt
Ca(s) Ca(s) + Cl+ Cl22(g)(g) → CaCl→ CaCl22(s)(s)
Reaction of magnesium oxide and carbon dioxide to Reaction of magnesium oxide and carbon dioxide to produce magnesium carbonateproduce magnesium carbonate
MgO(s) + COMgO(s) + CO22(g) →MgCO(g) →MgCO33(s)(s)
Decomposition ReactionsDecomposition Reactions
Reactions that produce two or more Reactions that produce two or more products from a single reactant.products from a single reactant.The general form for the reaction is The general form for the reaction is
AB AB → A + B→ A + B
Decomposition reactionsDecomposition reactions
Examples includeExamples include The removal of water from a hydrate (a substance The removal of water from a hydrate (a substance
that has water molecules linked in it’s structure)that has water molecules linked in it’s structure)
CuSOCuSO44·5H·5H22O(s) →O(s) →CuSOCuSO44(s) + (s) + 5H5H22O(g)O(g)
The heating of calcium carbonate to produce calcium The heating of calcium carbonate to produce calcium oxide and carbon dioxide gasoxide and carbon dioxide gas
CaCOCaCO33(s) → CaO(s) + CO(s) → CaO(s) + CO22(g)(g)
Replacement Reactions—Single Replacement Reactions—Single ReplacementReplacement
Single replacement reactions is where one Single replacement reactions is where one atom replaces another in the compoundatom replaces another in the compoundThe general formula is The general formula is
A + BC A + BC → AC + B→ AC + B
Replacement ReactionsReplacement Reactions
Examples includeExamples include
The replacement of copper by zinc in The replacement of copper by zinc in copper sulfate forming zinc sulfatecopper sulfate forming zinc sulfate
Zn(s) + CuSOZn(s) + CuSO44(aq) (aq) → Zn → Zn SOSO44(aq) + Cu(s)(aq) + Cu(s)
Replacement Reactions—Double Replacement Reactions—Double ReplacementReplacement
Two compounds that “switch” atoms with Two compounds that “switch” atoms with each othereach otherThe general formula isThe general formula is
AB + CD AB + CD → AD + CB→ AD + CB
Replacement Reactions—Double Replacement Reactions—Double ReplacementReplacement
Examples includeExamples includeThe formation of salt and water with the The formation of salt and water with the reaction of a base and an acidreaction of a base and an acid
HCl(aq) + NaOH(aq) HCl(aq) + NaOH(aq) →H→H22O(l) + NaCl(aq)O(l) + NaCl(aq)
Types of Chemical ReactionsTypes of Chemical Reactions
There are four main types of chemical There are four main types of chemical reactionsreactions Precipitation reactionsPrecipitation reactions
Reactions with OxygenReactions with Oxygen
Acid-base reactionsAcid-base reactions
Oxidation-reduction reactionsOxidation-reduction reactions
Precipitation reactionsPrecipitation reactions
A chemical change that produces an A chemical change that produces an insoluble product that will form a solid. insoluble product that will form a solid. Usually the solid can be seen “falling out” Usually the solid can be seen “falling out” of the solution, hence, called precipitation. of the solution, hence, called precipitation. At other times the solid makes the solution At other times the solid makes the solution turn from clear to cloudy.turn from clear to cloudy.
Solubility predictionsSolubility predictionsNa, K, and ammonium compounds are generally soluble.Na, K, and ammonium compounds are generally soluble.
Nitrates and acetates are generally solubleNitrates and acetates are generally soluble
Chlorides, bromides, and iodides are generally soluble. However, Chlorides, bromides, and iodides are generally soluble. However, iodine compounds that contain lead, silver, and mercury are iodine compounds that contain lead, silver, and mercury are insoluble.insoluble.
Carbonates and phosphates are generally insoluble. Sodium, Carbonates and phosphates are generally insoluble. Sodium, potassium, and ammonium carbonates and phosphates are soluble.potassium, and ammonium carbonates and phosphates are soluble.
Hydroxides and sulfides are generally insoluble. Sodium, Hydroxides and sulfides are generally insoluble. Sodium, potassium, calcium, and ammonium compounds are however potassium, calcium, and ammonium compounds are however soluble. soluble.
Reactions with oxygenReactions with oxygen
Many substances react with oxygen. If the Many substances react with oxygen. If the substance contains carbon, then carbon substance contains carbon, then carbon dioxide is usually produced. If the dioxide is usually produced. If the substance contains hydrogen, then water substance contains hydrogen, then water is usually produced.is usually produced.An example is iron turning to rustAn example is iron turning to rust
4Fe(s) + 3O4Fe(s) + 3O22(g) (g) → 2Fe→ 2Fe22OO33(s)(s)This number is called a coefficient, and indicates the
number of molecules or moles that reacts with the other compounds.
Acid-base ReactionsAcid-base ReactionsThis involves an acid combining with a This involves an acid combining with a base to form a salt.base to form a salt.An example would beAn example would be
HCl(aq) + NaOH(aq) HCl(aq) + NaOH(aq) → NaCl(aq) + H→ NaCl(aq) + H22O(aq)O(aq)
Oxidation Reduction ReactionsOxidation Reduction Reactions
Involves the transfer of negative charge Involves the transfer of negative charge from one reactant to another.from one reactant to another.The reaction of zinc with copper would be The reaction of zinc with copper would be an example.an example.
Zn(s) + CuZn(s) + Cu2+2+ (aq) (aq) → → ZnZn2+2+ (aq) + Cu (s) (aq) + Cu (s)
These specific reactions will be discuss in These specific reactions will be discuss in further detail at a later time.further detail at a later time.
Homework Assignment #12Homework Assignment #12
p. 147 Answer questions 59-68p. 147 Answer questions 59-68Read p. 126-136. Answer the exercises in Read p. 126-136. Answer the exercises in
the reading that were not answered in the reading that were not answered in class as examples.class as examples.
Chapter 4: Calculations Chapter 4: Calculations and the Chemical Equationand the Chemical Equation
Section 4: Balancing Chemical Section 4: Balancing Chemical EquationsEquations
The chemical equationThe chemical equation
Shows the molar quantity of reactants Shows the molar quantity of reactants needed to produce a certain molar needed to produce a certain molar quantity of products.quantity of products.
sincesince
The number of atoms in a molecule cannot The number of atoms in a molecule cannot be changed (it would make an entirely be changed (it would make an entirely different compound)different compound)
coefficientscoefficients
(whole numbers that show the numbers of (whole numbers that show the numbers of entire molecules) are used to balance a entire molecules) are used to balance a chemical equationchemical equation
For example, in the equationFor example, in the equation
CaCOCaCO33(s) (s) →→∆ ∆ CaO(s) + COCaO(s) + CO22(g)(g)
On the reactant sideOn the reactant side On the product sideOn the product side
1 mole Ca1 mole Ca 1 mole Ca1 mole Ca1 mole C1 mole C 1 mole C1 mole C3 moles of O3 moles of O 3 moles of O3 moles of O
Since there are the same numbers of each type of molecules on both sides of the arrow, the equation is balanced.
HCl(aq) + Ca(s) HCl(aq) + Ca(s) CaClCaCl22(s) + H(s) + H22(g)(g)
On the reactant sideOn the reactant side On the product sideOn the product side
1 mole H1 mole H 2 mole H2 mole H
1 mole Cl1 mole Cl 2 moles Cl2 moles Cl
1 mole Ca1 mole Ca 1 mole Ca1 mole Ca
Since there are not the same number of moles on both sides of this equation, the equation is not balanced.
HCl(aq) + Ca(s) HCl(aq) + Ca(s) CaClCaCl22(s) + H(s) + H22(g)(g)
2HCl(aq) + Ca(s) 2HCl(aq) + Ca(s) CaCl CaCl22(s) + H(s) + H22(g)(g)
To balance the equation, place a coefficient of 2 in front of the HCl.
Steps to balancing a chemical Steps to balancing a chemical equationequation
Step 1: Count the number of moles of Step 1: Count the number of moles of atoms of each element on both product atoms of each element on both product and reactant sideand reactant side
H2(g) + O2(g) H2O(l)
On the reactant side:
2 moles of H2 moles of O
On the product side:
2 moles of H1 mole of O
Steps to balancing a chemical Steps to balancing a chemical equationequation
Step 2: Determine which elements are not Step 2: Determine which elements are not balanced.balanced.
H2(g) + O2(g) H2O(l)
The oxygen atoms are not balanced in
this equation.
Steps to balancing a chemical Steps to balancing a chemical equationequation
Step 3: Balance one element at a Step 3: Balance one element at a time. time.
H2(g) + O2(g) H2O(l)
First: H2(g) + O2(g) 2H2O(l)
Then 2H2(g) + O2(g) 2H2O(l)
Steps to balancing a chemical Steps to balancing a chemical equationequation
Step 4: After you believe you have Step 4: After you believe you have successfully balanced the equation, check successfully balanced the equation, check to make sure you have the same number to make sure you have the same number of atoms on both sides of the equation.of atoms on both sides of the equation.
2H2(g) + O2(g) 2H2O(l)
4 moles of H2 moles of O
4 moles of H2 moles of O
Homework Assignment #13Homework Assignment #13
p. 147 Answer questions 69-84p. 147 Answer questions 69-84Read p. 136-145. Answer the exercises in Read p. 136-145. Answer the exercises in
the reading that were not answered in the reading that were not answered in class as examples.class as examples.
Chapter 4: Calculations Chapter 4: Calculations and the Chemical Equationand the Chemical Equation
Section 4.5: Calculations Using Section 4.5: Calculations Using the Chemical Equationthe Chemical Equation
Calculations Using Chemical Calculations Using Chemical EquationsEquations
Using the chemical formulas to calculate Using the chemical formulas to calculate amounts of materials needed or produced amounts of materials needed or produced can be done once you have a balanced can be done once you have a balanced chemical equation.chemical equation.
In order to carry out In order to carry out chemical calculations the chemical calculations the following guidelines must following guidelines must
be followed.be followed.The chemical formulas of ALL the The chemical formulas of ALL the products and reactants must be products and reactants must be knownknown
The basis for the calculations is the The basis for the calculations is the balanced chemical equation. Be balanced chemical equation. Be sure all of the equations are sure all of the equations are balanced first.balanced first.The calculations are performed in The calculations are performed in terms of moles.terms of moles.
Use of Conversion Factors--Use of Conversion Factors--conversion between moles and conversion between moles and
grams.grams.
Example: convert 10 moles of NaCl to gramsExample: convert 10 moles of NaCl to grams
The formula mass of NaCl is the molecular The formula mass of NaCl is the molecular mass of Na + the molecular mass of Cl mass of Na + the molecular mass of Cl (22.99 + 35.45 = 58.44 grams per mole.(22.99 + 35.45 = 58.44 grams per mole.
10 moles NaCl X 10 moles NaCl X 55.44 grams 55.44 grams = 554.4 grams of NaCl = 554.4 grams of NaCl1 mole1 mole
Use of Conversion Factors--conversion Use of Conversion Factors--conversion between moles and grams.between moles and grams.
Example: How many moles of CaClExample: How many moles of CaCl22 would would 23 grams contain?23 grams contain?
The formula mass of CaClThe formula mass of CaCl22 equals the equals the atomic mass of Ca and 2 X the atomic mass atomic mass of Ca and 2 X the atomic mass of Cl (40.08 + 2(35.45) = 110.98)of Cl (40.08 + 2(35.45) = 110.98)
23 grams X 23 grams X 1 mole1 mole = 0.207 moles = 0.207 moles 110.98 grams 110.98 grams
Use of Conversion Factors—Use of Conversion Factors—Conversion of moles of reactants Conversion of moles of reactants
to moles of products.to moles of products.
Once you have a balanced chemical Once you have a balanced chemical equation, develop a conversion equation, develop a conversion factor of reactants to products. After factor of reactants to products. After you have done that you can:you have done that you can: Calculate reacting quantitiesCalculate reacting quantities Calculate grams of product producedCalculate grams of product produced Relate the mass of reactants and Relate the mass of reactants and
productsproducts
Converting moles of reactants to Converting moles of reactants to moles of productmoles of product
2H2H22(g) + O(g) + O22(g) (g) 2H 2H22O(l)O(l)
In this equation: 2 moles of HIn this equation: 2 moles of H2 2 will react with will react with 1 mole of O1 mole of O22 to produce 2 moles of H to produce 2 moles of H22O.O.
2H2H22(g) + O(g) + O22(g) (g) 2H 2H22O(l)O(l)
To calculate the number of grams of HTo calculate the number of grams of H22O O produced by 1 mole of Oproduced by 1 mole of O22
1.1. Convert from moles of OConvert from moles of O2 2 to moles of Hto moles of H22OO
1 mole O2 x 2 moles H2O = 2 moles H2O produced
2. Convert the moles of H2. Convert the moles of H22O to grams of O to grams of HH22OO
2 moles X 18.016 grams = 1 mole36.032 grams
Relating masses of reactants Relating masses of reactants and productsand products
CaCOCaCO33(s) (s) →→∆ ∆ CaO(s) + COCaO(s) + CO22(g)(g)
How many grams of Ca0 will be produced How many grams of Ca0 will be produced by 100.0 grams of by 100.0 grams of CaCOCaCO33??
First, determine how many moles First, determine how many moles will be produced.will be produced.
CaCOCaCO33(s) (s) →→∆ ∆ CaO(s) + COCaO(s) + CO22(g)(g)
One mole of CaCO3 will produce one mole of
CaO
Next, determine how many grams of Next, determine how many grams of each are in each mole of substance.each are in each mole of substance.
CaCO3(s) →∆ CaO(s) + CO2(g)
1 mole of CaCO3 has a formula mass
of 100.09 grams
1 mole of CaO has a
mass of 56.06 grams
0.9991 moles CaCO3 will produce 0.9991 moles CaO0.9991 moles CaCO3 will produce 0.9991 moles CaO
0.9991 moles CaO X 0.9991 moles CaO X 56.06 grams56.06 grams = 56.00 grams = 56.00 grams CaOCaO
1 mole1 mole
100 grams CaCO3 X 1 mole CaCO3 = 0.9991 moles CaCO3
100.09 grams CaCO3
Theoretical and percent yieldTheoretical and percent yield
If a chemical reaction occurs, in theory you If a chemical reaction occurs, in theory you can calculate how much of the product is can calculate how much of the product is created. This would be the maximum created. This would be the maximum amount that is produced. However, in the amount that is produced. However, in the real world often not all the possible real world often not all the possible product are produced in a chemical product are produced in a chemical reaction.reaction.
Theoretical yieldTheoretical yield
The maximum amount of product that The maximum amount of product that could be produced determined by could be produced determined by calculations using the chemical equation.calculations using the chemical equation.
Percent yieldPercent yield
The ratio of the actual and theoretical The ratio of the actual and theoretical yields determined by the formulayields determined by the formula
%yield = %yield = actual yieldactual yield X 100% X 100%Theoretical yieldTheoretical yield
Example:Example:
2HCl(aq) + Ca(s) 2HCl(aq) + Ca(s) CaCl CaCl22(s) + H(s) + H22(g)(g)
Assume the theoretical yield of CaClAssume the theoretical yield of CaCl2 2 in in this equation were 30 g. If the actual this equation were 30 g. If the actual yield of CaClyield of CaCl2 2 were 25 g, calculate the were 25 g, calculate the percentage yield.percentage yield.
%yield = %yield = actual yieldactual yield X 100% X 100%Theoretical yieldTheoretical yield
25 g 25 g X 100 = 83.3% X 100 = 83.3% 30g30g
Homework AssignmentHomework Assignment
p. 147-148 Exercises 85-104 (odd)p. 147-148 Exercises 85-104 (odd)